Consider the real number $\sqrt{2}+\sqrt{3}$.
When we calculate the even powers of $\sqrt{2}+\sqrt{3}$ we get:
$(\sqrt{2}+\sqrt{3}{)}^2=9.898979485566356\cdots$
$(\sqrt{2}+\sqrt{3}{)}^4=97.98979485566356\cdots$
$(\sqrt{2}+\sqrt{3}{)}^6=969.998969071069263\cdots$
$(\sqrt{2}+\sqrt{3}{)}^8=9601.99989585502907\cdots$
$(\sqrt{2}+\sqrt{3}{)}^{10}=95049.999989479221\cdots$
$(\sqrt{2}+\sqrt{3}{)}^{12}=940897.9999989371855\cdots$
$(\sqrt{2}+\sqrt{3}{)}^{14}=9313929.99999989263\cdots$
$(\sqrt{2}+\sqrt{3}{)}^{16}=92198401.99999998915\cdots$

It looks as if the number of consecutive nines at the beginning of the fractional part of these powers is non-decreasing.
In fact it can be proven that the fractional part of $(\sqrt{2}+\sqrt{3}{)}^{2n}$ approaches $1$ for large $n$.

Consider all real numbers of the form $\sqrt{p}+\sqrt{q}$ with $p$ and $q$ positive integers and $p<q$, such that the fractional part of $(\sqrt{p}+\sqrt{q}{)}^{2n}$ approaches $1$ for large $n$.

Let $C(p,q,n)$ be the number of consecutive nines at the beginning of the fractional part of $(\sqrt{p}+\sqrt{q}{)}^{2n}$.

Let $N(p,q)$ be the minimal value of $n$ such that $C(p,q,n)\ge 2011$.

Find ${\displaystyle \sum N(p,q)\text{ for }p+q\le 2011}$.